Thread: Not sure what kind of loop I need to use for an intersection program

We were given a code that we need to expand from. The code finds and prints the x and y coordinates of the intersection point of a line and it's curve. We have to modify and add to the program so that it finds all the coordinates where x varies from 0.9 to 2.0 with 0.1 increments. It also has to calculate the number of steps taken to find the solution.

// Homework Program Intersection inter.java
 
public class inter
 
{
	public static void main(String[] args)
	{
		double x;
		double stepsize;
		int counter;
 
		{
		x = 0.0;
		stepsize = 1.0;
		counter = 0;
 
		{
		while (stepsize >= 0.01)
		{
			while ((x*x - 4.0*x + 3.7) > (0.9*x))
			{
				x = x + stepsize;
			}
			x = x-stepsize;
			stepsize = stepsize / 10;
 
			if (stepsize <=2)
				{
				++counter;
				}
 
		}
		System.out.printf("%s%6.2f\n","The line 0.9X and the curve meet at x = ",x);
		System.out.printf("%s%6.2f\n", " and y =", x*0.9);
	}
}
} // end of main method
} // end of the class

I tried to add in

for ( double count = 0.9; count <= 2; count+=.1)

but it just repeated the initial thing over and over again. I'm not sure if I should use an IF loop to make this work. The counter isn't adding correctly either, but that's not surprising since it's not doing the other part right either. We aren't on arrays yet.

I really am not grasping java programming. This is due tomorrow night. I was in a car accident on Tuesday that has thrown me way off track this week. If anybody is willing to help give me an idea of where to go with this, I'd really appreciate it. I'm not looking to have it done for me, just give me an idea of what kind of loop will make this work? I think I can probably get it from there.

Originally Posted by sessypeanut

We were given a code that we need to expand from. The code finds and prints the x and y coordinates of the intersection point of a line and it's curve. We have to modify and add to the program so that it finds all the coordinates where x varies from 0.9 to 2.0 with 0.1 increments.

So, the problem is: A line is determined by the equation y = a * x, where you are given a = 0.9.

Find approximate values of x and y for which the line intersects the curve determined by the equation y = x*x - 4x + 3.7

Your program does this and gives (x,y) = (0.93, 0.84) with an error in x of something less than 0.01.

I think the assignment might be to let the value of a go from 0.9 through 2.0 in steps of 0.1 and find an approximate intersection point for each such line with the given curve.

I'll rearrange your original program slightly in anticipation of letting the line be defined by a*x (where a is a variable) rather than being hardcoded to 0.9*x:

public class Z {
    public static void main(String [] args) {
        double x;
        double stepsize;
        double a;
 
        a = 0.9;
 
        x = 0.0;
        stepsize = 1.0;
        while (stepsize >= 0.01) {
            while ((x * x - 4.0 * x + 3.7) > (a * x)) {
                x = x + stepsize;
            }
            x = x - stepsize; 
            stepsize = stepsize / 10;
        }
        System.out.printf("The line %.1fX and the curve intersect at approximately (%.2f,%.2f)\n",
                a, x, a*x);
    } // end of main method
} // end of the class

I left out the counter stuff since it makes absolutely no sense to me.

Anyhow, the output is

The line 0.9X and the curve intersect at approximately (0.93,0.84)


Now, just make a loop that lets a go from 0.9 through 2.0 and put the calculations (starting with x = 0) and printout inside that loop

As for the requirement to count the number of steps, that still makes no sense to me, so maybe you can ask your instructor.

Bottom line:

Sometimes it helps to visualize what the program is actually doing. I have attached a plot of the curve and the lines for values of a that I think you need. Each line intersects the parabola at two points. For a given value of a, the program finds an approximate value of x for the intersection closest to the origin.

Final comment: If that is not what your assignment is, well, maybe you can articulate it in a way that we can understand. I gave it a shot.

Cheers!

Z

How do you find the second intersection?

Originally Posted by Zaphod_b

So, the problem is: A line is determined by the equation y = a * x, where you are given a = 0.9.

Find approximate values of x and y for which the line intersects the curve determined by the equation y = x*x - 4x + 3.7

Your program does this and gives (x,y) = (0.93, 0.84) with an error in x of something less than 0.01.

I think the assignment might be to let the value of a go from 0.9 through 2.0 in steps of 0.1 and find an approximate intersection point for each such line with the given curve.

I'll rearrange your original program slightly in anticipation of letting the line be defined by a*x (where a is a variable) rather than being hardcoded to 0.9*x:

public class Z {
    public static void main(String [] args) {
        double x;
        double stepsize;
        double a;
 
        a = 0.9;
 
        x = 0.0;
        stepsize = 1.0;
        while (stepsize >= 0.01) {
            while ((x * x - 4.0 * x + 3.7) > (a * x)) {
                x = x + stepsize;
            }
            x = x - stepsize; 
            stepsize = stepsize / 10;
        }
        System.out.printf("The line %.1fX and the curve intersect at approximately (%.2f,%.2f)\n",
                a, x, a*x);
    } // end of main method
} // end of the class

I left out the counter stuff since it makes absolutely no sense to me.

Anyhow, the output is

The line 0.9X and the curve intersect at approximately (0.93,0.84)


Now, just make a loop that lets a go from 0.9 through 2.0 and put the calculations (starting with x = 0) and printout inside that loop

As for the requirement to count the number of steps, that still makes no sense to me, so maybe you can ask your instructor.

Bottom line:

Sometimes it helps to visualize what the program is actually doing. I have attached a plot of the curve and the lines for values of a that I think you need. Each line intersects the parabola at two points. For a given value of a, the program finds an approximate value of x for the intersection closest to the origin.

Final comment: If that is not what your assignment is, well, maybe you can articulate it in a way that we can understand. I gave it a shot.

Cheers!

Z

Originally Posted by lucky13

How do you find the second intersection?

Here's a recap of how this method found the the first intersection: (Refer to the graphs in my previous post.)

For a given value of a we know the the point on the curve at x = 0 is above the the point on the line at that value of x. (That is, evaluating the equation of curve at x = 0 gives a number that is greater than the value obtained by evaluating the equation for the line at x = 0.)

So: Find an approximation of the value of x where the two intersect by incrementing x until we find a value of x for which the point on the curve is below the point on the line.

Then, we know that the intersection is between the last two points. This can be proved since function for the curve and the function for the line are both continuous. (Refer to the Intermediate Value Theorem in your favorite elementary mathematical analysis text.)

Now, to find the second intersection, start at the value of x for which the point on the curve is below the point on the line. That is, start with a point just to the right of the approximate value of x of the first intersection. The point is below the line, right? Then increment x until we find a value of x for which the point on the curve is above the point on the line. Then we know that there is an intersection between the last two values of x.

In other words, add another set of nested loops after the loops in the previous program. The second set of loops looks pretty much like the first, but the termination condition of the inner loop has '<' rather than '>' in the expression.

For a given value of a, the result from the enhanced program could look something like

The line 0.9X and the curve intersect at approximately (0.93,0.84)
The line 0.9X and the curve intersect at approximately (3.96,3.56)

Also...

I now realize what was meant in the assignment by counting the number of steps taken. I'm not sure why I didn't see it before:

Define an int variable named counter.

Before the outer loop, set counter equal to zero.
Increment the counter value every time it goes through the inner loop. (That is, increment the variable each time you get a new value of x.)
Print the value of counter after the outer loop terminates.

Bottom line: If you can find two values of x where the line is above the curve for one value and the line is below the curve for the second value, you know they intersect somewhere between those two values of x. For any given functions, looking at their graphs might give some insight into developing a search strategy.

The assignment defines a strategy that can work for the given functions, starting at x = 0 and with an initial step size of 1. These values might not work for other functions.


Cheers!

Z

Thank you very much for the help.

I now understand.

Best Regards

Originally Posted by Zaphod_b

Here's a recap of how this method found the the first intersection: (Refer to the graphs in my previous post.)

For a given value of a we know the the point on the curve at x = 0 is above the the point on the line at that value of x. (That is, evaluating the equation of curve at x = 0 gives a number that is greater than the value obtained by evaluating the equation for the line at x = 0.)

So: Find an approximation of the value of x where the two intersect by incrementing x until we find a value of x for which the point on the curve is below the point on the line.

Then, we know that the intersection is between the last two points. This can be proved since function for the curve and the function for the line are both continuous. (Refer to the Intermediate Value Theorem in your favorite elementary mathematical analysis text.)

Now, to find the second intersection, start at the value of x for which the point on the curve is below the point on the line. That is, start with a point just to the right of the approximate value of x of the first intersection. The point is below the line, right? Then increment x until we find a value of x for which the point on the curve is above the point on the line. Then we know that there is an intersection between the last two values of x.

In other words, add another set of nested loops after the loops in the previous program. The second set of loops looks pretty much like the first, but the termination condition of the inner loop has '<' rather than '>' in the expression.

For a given value of a, the result from the enhanced program could look something like

The line 0.9X and the curve intersect at approximately (0.93,0.84)
The line 0.9X and the curve intersect at approximately (3.96,3.56)

Also...

I now realize what was meant in the assignment by counting the number of steps taken. I'm not sure why I didn't see it before:

Define an int variable named counter.

Before the outer loop, set counter equal to zero.
Increment the counter value every time it goes through the inner loop. (That is, increment the variable each time you get a new value of x.)
Print the value of counter after the outer loop terminates.

Bottom line: If you can find two values of x where the line is above the curve for one value and the line is below the curve for the second value, you know they intersect somewhere between those two values of x. For any given functions, looking at their graphs might give some insight into developing a search strategy.

The assignment defines a strategy that can work for the given functions, starting at x = 0 and with an initial step size of 1. These values might not work for other functions.


Cheers!

Z